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The accepted winner is isaacg, with his 7-bit ASCII answer. However, the challenge isn't over yet - this bounty is awarded to the shortest answer. If, somehow, you get all the first characters of all the other answers down into 10 bytes, you will win the bounty. This includes all the characters from Round 2's GolfScript answer (plus the one added by that answer itself). This is the ONE time I will let you go out of order - if you have any objections to this, let me know in the comments.

I'd like to give credit to randomra, who helped me with my old idea and gave me this new one.

Previous Winners

  • Round 1: isaacg, with 7-bit ASCII
    Next bytes: 30 (or 10 if you want that sweet, sweet rep)
    You know, code-golfing is really cool. People take a challenge, and it slowly gets smaller! But let's do this another way. So, here is my challenge:

  • The code will print the first character of all the previous answers in the order they were posted (the first answer prints nothing)

  • The code starts at 100 bytes and decreases by 5 each time.
  • If two posts are going by the same answer (i.e, they both posted within a few seconds of each other), the newer one has to add the old one's character and decrease by 5 bytes (even by a few seconds).
  • Any language can be used.
  • Your code must not produce any errors.
  • Your code must use all the bytes required for the first step.
  • Your code must print to STDOUT.
  • Non-printable characters are OK, but:
    • They can not be the first character (for the sake of the purpose of this question)
    • You must let everyone know where they are
  • You may post multiple answers, but:
    • You must wait 2 answers before posting another (so if you posted the 100 bytes, you have to wait until 85 bytes.)
  • You can't:
    • use more than 10 bytes of comments
    • have variables that go unused for the entire program
    • fill the program with whitespace
    • have variable names longer than 10 bytes (but you can have multiple less-than 10-byte variables)
      (EMBLEM's first answer being the exception to these rules, because it was posted before these restrictions.)
  • No standard loopholes. Unless you want to take all the fun out of the challenge.
  • When no more answers are submitted for 3 weeks, the answer that uses the fewest bytes wins. (In the event of a tie, the one printing the longer string wins.)

Example: The third answer has to be a 90 bytes long code outputting two characters (the first char of the 100-byte code then the first char of the 95-byte code). The first answer outputs nothing (no previous answers to get chars from).

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  • \$\begingroup\$ "Unnecessary whitespace" is bad wording. Python's whitespace to end statements is unnecessary, because you can use semicolons. \$\endgroup\$ – EMBLEM Apr 7 '15 at 18:17
  • 8
    \$\begingroup\$ Perhaps, we can start at 100 bytes again, but start with ppuppPq([#fwSmdP[ as the starting string, and restarting each time the string is longer than the allowed bytecount. I can't think of a winning criterion then, though. \$\endgroup\$ – Sanchises Apr 7 '15 at 20:22
  • 1
    \$\begingroup\$ @Scimonster You honestly think there is a 4-bit character set that includes all of #(PS[dfmpquw? We're running into the very limits of information density here. Unless you write a 10-byte program that processes all previous answers. Not sure if that's worth waiting for. \$\endgroup\$ – Sanchises Apr 7 '15 at 21:30
  • 2
    \$\begingroup\$ @sanchises We're not saying there is one. We're giving people a chance to see what they can do. If no one can, we'll restart it. \$\endgroup\$ – ASCIIThenANSI Apr 7 '15 at 21:35
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    \$\begingroup\$ -1 Why didn't this challenge just die with dignity after the 15-byte solution? It seems disingenuous to change it after a good, "winning" answer has been given just so you can keep playing by different rules. \$\endgroup\$ – Geobits Apr 9 '15 at 13:07

32 Answers 32

0
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Round 2: VBS, 80 bytes

s = "ppuppPq([#fwSmdP[ppnvo" : For i=1 To 22 : WScript.Echo Mid(s ,i , 1) : Next

Left field language.

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Round 2: Python, 85 bytes

os=['p']*8
os[4:6]=u'P'+r'q([#fwSmdP'
os[2:2]='u'
os[-2:-2]='['
print''.join(os)+'nv'

Starts with the fundamentals - the 'p's - and builds it up from there.

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  • \$\begingroup\$ Prints ppuppPq([#fwSmdP[ppp, should be ppuppPq([#fwSmdP[ppnv \$\endgroup\$ – EMBLEM Apr 9 '15 at 2:32
  • \$\begingroup\$ @EMBLEM: Oops, silly me. Thanks, fixed. \$\endgroup\$ – Claudiu Apr 9 '15 at 5:08

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